Five positive integers
A, B, C, D and E, with
A < B < C < D < E, are such that:
(i)
A, B and C (in this order) are in arithmetic sequence, and:
(ii)
B, C and D (in his order) are in geometric sequence, and:
(iii)
C, D and E (in this order) are in
harmonic sequence.
Determine the
minimum value of
(EA) such that there are
precisely two quintuplets (A, B, C, D, E) that satisfy all the given conditions.
Note: Try to solve this problem analytically, although computer program/ spreadsheet solutions are welcome.
(In reply to
re: computer solutions  no proof by Dej Mar)
"Using substitution I was able to derive the equation,
[EA] = (4/A)*x^{2} + 4*x,
such that x is the common difference between A and B. "
The above can be used to show that no further solutions exist with EA at or below 64.
As x is the difference between B and A (the common difference of either BA or CB), and that difference must also be less than 64 in order for EA to be less than or equal to 64, for a solution 4*x^2 must be less than 4*64^2 or 16,384.
But as that is the case, in order for EA to be an integer, A must be less than 16,384. Since I checked all the values up through 200,000, all the possibilities are covered for all EA of less than or equal to 64, and 64 is the answer.

Posted by Charlie
on 20081029 11:58:22 